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A point moves along a curved path y=0.4x2, At x=2m its speed is 6 m/s increasing at 3 m/s2. At this instatnt find- (i) Velocity components along x and y direction. (ii) Its acceleration.
written 3.9 years ago by gravatar for teamques10 teamques10 ★ 69k •   modified 3.9 years ago
engineering physics
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written 3.9 years ago by gravatar for teamques10 teamques10 ★ 69k

$y= 0.4\ x^2 \cdots (Given)\ Hence,\ \dfrac {dy} {dx}= 0.8x\ and \dfrac {d^2y} {dx^2}=0.8\ At\ x=2, \dfrac {dy} {dx}= 0.8 \times 2= 1.6\ and \dfrac {d^2y} {dx^2}=0.8\$

$Radius\ of\ curvature\ (\rho)= {\bigg [1+ \ (dy/dx)^2 \bigg] ^{3/2} \over{d^2y / dx^2}}\\ {\bigg [1+ (1.6)^2 \bigg] ^{3/2} \over{0.8}}\\ = 8.3962 \ m. …

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